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                             Strength of Reinforced Concrete  - Doubly Reinforced Rectangular Section  

Problem 9-2

Compute the nominal flexural strength Mn of the reinforced concrete rectangular section given below in figure 9-2(a).  Take fc′ = 5000 psi,      width b = 13 in., effective depth d = 25 in. The beam has tension steel of As = 8-#9 bars placed in two layers with a spacing of 1 in. between them, fy = 40,000 psi, and compression steel of  As′ = 2-#8 bars, fs′ = 40,000 psi, effective cover  d′ of 3 in.

doubly reinforced concrete beam

Figure 9-2

Solution:

The given section is doubly reinforced with steel in tension as well as compression zone of the section. The computation of nominal flexural strength Mn is based on the guidelines of ACI-318. The maximum value of usable strain at the extreme concrete  fiber is assumed  to be 0.003.In the case of two layers of tension steel, the effective depth is taken from the centroid of tension steel to the top edge of the section.

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For  fc′  grerater than 4000 psi the value  of β1 is calculated as given below;

 β1  = 0.85 - 0.05 {( fc′ -4000)/1000} = 0.8

Assume that the compression steel has yielded when the strength is reached (strain in concrete is 0.003).

Given that tension steel consists of 8 bars of #9 (dia 1.128 in.) .

       Area of one bar of #9 = 1 in2.

       As = 8-#9 bars. = 8 (1.00) = 8 in2.

     Area of compression steel , As′ = 2-#8 = 2(0.79) = 1.58 in2.

The internal forces acting on the section  shown in figure 9-2(c) are calculated as given below;

   Cc = 0.85 fcba = 0.85 (5) (13) a = 55.25 a

   Cs = (fs′ - 0.85 fc′) As′ = (40 - 0.85*5) 1.58 = 56.485 kips

   T =  As fy  =  (8) (40) = 320 kips

 Applying static equilibrium, we get  Cc + Cs = T;

                  55.25 a + 56.485 = 320        

therefore depth of stress block,  a = (320 - 56.485)/55.25 = 4.77 in

Depth of neutral axis   x = a / β1  = 4.77/0.8 = 5.96 in

             Cc = 55.25 (4.77) = 263.54 kips

 

By straight line proportion (figure 9-2(b)) we can calculate the strain in compression steel when the extreme concrete fiber has a compressive strain of 0.003.

     εs′ = ( x - d′ ) (0.003) / x = (5.96 - 3) (0.003) / 5.96 =  0.00149

      yield strain of steel, εy = fy /Es = 40 / 29000 = 0.00138

     εs′  is greater than  εy , this means that compression steel has yielded before crushing of concrete, hence the assumption is verified and the value of x is valid.

If Compression steel does not yield; then ?? see Example 9-3

Now we determine the strain in the two layers of tension steel.

Using strain diagram we can calculate the strain in tension steel

  Lower most layer of tension steel

    εs1 = 0.003(25 + 0.5 + 1.28/2 - 5.96) / 5.96 = 0.01

  Second layer of tension steel

    εs2 = 0.003(25 - 0.5 - 1.28/2 - 5.96) / 5.96 = 0.009

  The strain in both the layers of tension steel is more than yield strain and also greater than 0.005. Hence the section is tension-controlled.

  The nominal flexural strength is computed as given below;

   Mn = Cc (d - a/2)  + Cs (d - d′ )

      =  263.54 (25 - 4.77/2) + 56.485 (25 - 3) = 7202.63 in-kips

    Mn = 600.22 ft-kips 

   see another problem 9-1 and problem 9-3 on Reinforced Concrete

You can also use our  Reinforced concrete calculator
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Last updated on Tuesday September 17, 2013